Suppose $ABC$ is a scalene right triangle, and $P$ is the point on   hypotenuse $\overline{AC}$ such that $\angle{ABP} =
45^{\circ}$. Given that $AP = 1$ and $CP = 2$, compute the area of   $ABC$.
Explanation: [asy]
pair A,B,C,P;

B = (0,0);
A = (0,1);
C = (2,0);
P = A + (C-A)/3;
draw(P--B--A--C--B);
label("$B$",B,SW);
label("$A$",A,NW);
label("$C$",C,SE);
label("$P$",P,NE);
[/asy]


Notice that $\overline{BP}$ bisects the right angle at $B$. Thus, the Angle Bisector Theorem tells us that $AB/BC = AP/PC = 1/2$. So, we have $AB = x$ and $BC = 2x$ for some $x$.  By the Pythagorean theorem, we have $5x^2 =AC^2 = 9$, so $x^2 = \frac95$.  Finally, the desired area $\frac{1}{2}(x)(2x) = x^2 = \boxed{\frac{9}{5}}$.